the answer is (m,n)=(1,1) i think it was hard for problem 1 but i solved all of qestions in the exam.the exam was harder than yesterday.

Consider gcd(n, m) = d --> n = ad, m = bd and gcd(a, b) = 1. Then the logic is the same when solving, for example, x^y = y^x in positive integers. These problems are really typical, but the special thing in this one is the fact that the exponent of n is (n^n).

First of all, it's obvious that m and n have same prime divisors. Take: m=((p∨1)^(a∨1))×((p∨2)^(a∨2))×.....× ((p∨k)^(a∨k)) (m>=2) n=((p∨1)^(b∨1))×((p∨2)^(b∨2))×.....× ((p∨k)^(b∨k)) (n>=2) we want to prove that : m<n^n. suppose that m>=n^n. then we have: n^(n^n) = m^m >= m^(n^n) >= (n^n)^(n^n) so: n>=n^n contradiction so we have: m < n^n ⇒ m^m < n^(mn) ⇒ n^(n^n) < n^(mn) ⇒ m > n^(n-1) now note that by counting the power of prime divisors in n^(n^n) and m^m we get that: (n^n)×(b∨i)=m×(a∨i) (0<i<k+1) so we have: (n^n)/(m) = (a∨i)×(b∨i) (0<i<k+1) now we have two cases: 1)m divides n^n. so (n^n)/(m) is an integer. but we know that m>n^(n-1) so there exist j such that (a∨j)>(n-1)×(b∨j) ⇒ (a∨j)/(b∨j)>(n-1) but (n^n)/(m) is an integer so (n^n)/(m) = (a∨j)×(b∨j) >=n so n^n>=mn contradiction. 2)m does not divide n^n. then there exist j such that (a∨j)>(n)×(b∨j) so (n^n)/(m) = (a∨j)×(b∨j)>n ⇒ n^n>mn ⇒ n^(n-1)>m contradiction so m and n do not have any prime divisors and they are both equal to 1.

Hint:take $m=n^a$ where $a$ is rational.

see here : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=466&t=150684